Yes, such an $O$ exists. In fact, one can construct $O$ so that
$\mathbb{P}\!\left(\sup_m s_O(x,m)=+\infty\right)=1$.
Here is one explicit construction.
For each integer $j\ge 1$, define
$k_j:=\lceil \log_2 j\rceil$.
Then $k_j\to\infty$ as $j\to\infty$, but it grows slowly enough that
$\sum_{j=1}^\infty 2^{-k_j}=\infty$. Next choose integers $n_j$ so that the blocks of coordinates we use are disjoint. For example, set
$n_1:=0$
and $n_{j+1}:=n_j+k_j+1$.Then the intervals $$I_j:=\{n_j+1,\dots,n_j+k_j\}$$ are pairwise disjoint. Now define the map $O:{0,1}^*\to{0,1}^*$ as follows. If a finite binary string $m$ has length $|m|=n_j$ for some $j$, then let
$O(m):=m\,0^{k_j}$,
that is, append $k_j$ zeros to the end of $m$. For all other strings $m$, define $O(m):=m$.
We now show that this choice works. Let $x=x_1x_2x_3\cdots$ be a random infinite binary sequence, with independent fair bits. For each $j$, consider the event
$E_j:=\{x_{n_j+1}=\cdots=x_{n_j+k_j}=0\}$.
This is the event that the block of bits of $x$ on the interval $I_j$ consists entirely of zeros. Since the intervals $I_j$ are pairwise disjoint, the events $E_j$ are independent. Also, because each bit is $0$ with probability $1/2$, we have
$\mathbb{P}(E_j)=2^{-k_j}$.
Now suppose that $E_j$ occurs. Let $m$ be the prefix of $x$ of length $n_j$, namely
$m=x_1x_2\cdots x_{n_j}$.
By construction,
$O(m)=m\,0^{k_j}$.
Because $E_j$ says that the next $k_j$ bits of $x$ are all zero, it follows that $O(m)$ is a prefix of $x$.
Therefore, from the definition of $s_O(x,m)$, we get:
$$s_O(x,m)=|O(m)|-|m|=(n_j+k_j)-n_j=k_j$$
So whenever $E_j$ occurs, there exists some $m$ such that
$s_O(x,m)=k_j$.
Hence $$\sup_m s_O(x,m)\ge k_j$$ on the event $E_j$. Now $\sum_{j=1}^\infty \mathbb{P}(E_j)=\sum_{j=1}^\infty 2^{-k_j}=\infty$,
and the events $E_j$ are independent. Therefore, by the second Borel--Cantelli lemma, with probability $1$, infinitely many of the events $E_j$ occur. But whenever $E_j$ occurs, we have
$\sup_m s_O(x,m)\ge k_j$,
and since $k_j\to\infty$, these lower bounds become arbitrarily large. Therefore, with probability $1$,
$\sup_m s_O(x,m)=+\infty$. So the answer to your question is yes, and in fact one can achieve the stronger statement
$$\boxed{\mathbb{P}\!\left(\sup_m s_O(x,m)=+\infty\right)=1.}$$
EDIT: Only one part needs to be corrected. The construction of $O$ stays the same, and the Borel--Cantelli argument stays the same. The only point that needs to be replaced is the step where one justifies that the modified score still goes to $+\infty$ after subtracting the penalty term $\ln|m|$. The correct statement to insert is the following. On the event $E_j$, if $m$ is the prefix of $x$ of length $n_j$, then:
$$s_O(x,m)=k_j-\ln n_j$$
Therefore it is enough to prove that
$k_j-\ln n_j\to+\infty$. Now
$n_j=1+\sum_{i=1}^{j-1}(k_i+1)$,
and since
$k_i=\lceil \log_2 i\rceil\le \log_2 i+1$,
we get
$n_j\le 1+\sum_{i=1}^{j-1}(\log_2 i+2)$. In particular, there exists a constant $C>0$ such that for all sufficiently large $j$,
$n_j\le C_j\log j$. Taking natural logarithms gives
$\ln n_j\le \ln C+\ln j+\ln\ln j$. On the other hand,
$k_j=\lceil \log_2 j\rceil\ge \log_2 j=\dfrac{\ln j}{\ln 2}$. Hence:
$$k_j-\ln n_j
\ge
\dfrac{\ln j}{\ln 2}-(\ln C+\ln j+\ln\ln j)$$ Equivalently,
$$k_j-\ln n_j\ge\left(\dfrac{1}{\ln 2}-1\right)\ln j-\ln\ln j-\ln C$$ Since $\dfrac{1}{\ln 2}-1>0$ and $\ln\ln j=o(\ln j)$, the right-hand side tends to $+\infty$. Therefore
$k_j-\ln n_j\to+\infty$.
